Optimal. Leaf size=77 \[ -\frac{\text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}+\frac{\text{sech}^{-1}\left (c e^{a+b x}\right )^2}{2 b}-\frac{\text{sech}^{-1}\left (c e^{a+b x}\right ) \log \left (e^{2 \text{sech}^{-1}\left (c e^{a+b x}\right )}+1\right )}{b} \]
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Rubi [A] time = 0.101172, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {2282, 6281, 5660, 3718, 2190, 2279, 2391} \[ -\frac{\text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}+\frac{\text{sech}^{-1}\left (c e^{a+b x}\right )^2}{2 b}-\frac{\text{sech}^{-1}\left (c e^{a+b x}\right ) \log \left (e^{2 \text{sech}^{-1}\left (c e^{a+b x}\right )}+1\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 6281
Rule 5660
Rule 3718
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \text{sech}^{-1}\left (c e^{a+b x}\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\text{sech}^{-1}(c x)}{x} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\cosh ^{-1}\left (\frac{x}{c}\right )}{x} \, dx,x,e^{-a-b x}\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int x \tanh (x) \, dx,x,\cosh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )\right )}{b}\\ &=\frac{\cosh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )^2}{2 b}-\frac{2 \operatorname{Subst}\left (\int \frac{e^{2 x} x}{1+e^{2 x}} \, dx,x,\cosh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )\right )}{b}\\ &=\frac{\cosh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )^2}{2 b}-\frac{\cosh ^{-1}\left (\frac{e^{-a-b x}}{c}\right ) \log \left (1+e^{2 \cosh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )}\right )}{b}+\frac{\operatorname{Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\cosh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )\right )}{b}\\ &=\frac{\cosh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )^2}{2 b}-\frac{\cosh ^{-1}\left (\frac{e^{-a-b x}}{c}\right ) \log \left (1+e^{2 \cosh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )}\right )}{b}+\frac{\operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \cosh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )}\right )}{2 b}\\ &=\frac{\cosh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )^2}{2 b}-\frac{\cosh ^{-1}\left (\frac{e^{-a-b x}}{c}\right ) \log \left (1+e^{2 \cosh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )}\right )}{b}-\frac{\text{Li}_2\left (-e^{2 \cosh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )}\right )}{2 b}\\ \end{align*}
Mathematica [B] time = 1.17138, size = 249, normalized size = 3.23 \[ x \text{sech}^{-1}\left (c e^{a+b x}\right )-\frac{\sqrt{\frac{1-c e^{a+b x}}{c e^{a+b x}+1}} \sqrt{c e^{a+b x}+1} \left (4 \text{PolyLog}\left (2,\frac{1}{2} \left (1-\sqrt{1-c^2 e^{2 (a+b x)}}\right )\right )-\log ^2\left (c^2 e^{2 (a+b x)}\right )-2 \log ^2\left (\frac{1}{2} \left (\sqrt{1-c^2 e^{2 (a+b x)}}+1\right )\right )+4 \log \left (\frac{1}{2} \left (\sqrt{1-c^2 e^{2 (a+b x)}}+1\right )\right ) \log \left (c^2 e^{2 (a+b x)}\right )+\left (8 b x-4 \log \left (c^2 e^{2 (a+b x)}\right )\right ) \tanh ^{-1}\left (\sqrt{1-c^2 e^{2 (a+b x)}}\right )\right )}{8 b \sqrt{1-c e^{a+b x}}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.361, size = 140, normalized size = 1.8 \begin{align*}{\frac{ \left ({\rm arcsech} \left (c{{\rm e}^{bx+a}}\right ) \right ) ^{2}}{2\,b}}-{\frac{{\rm arcsech} \left (c{{\rm e}^{bx+a}}\right )}{b}\ln \left ( 1+ \left ({\frac{1}{c{{\rm e}^{bx+a}}}}+\sqrt{{\frac{1}{c{{\rm e}^{bx+a}}}}-1}\sqrt{{\frac{1}{c{{\rm e}^{bx+a}}}}+1} \right ) ^{2} \right ) }-{\frac{1}{2\,b}{\it polylog} \left ( 2,- \left ({\frac{1}{c{{\rm e}^{bx+a}}}}+\sqrt{{\frac{1}{c{{\rm e}^{bx+a}}}}-1}\sqrt{{\frac{1}{c{{\rm e}^{bx+a}}}}+1} \right ) ^{2} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} b c^{2} \int \frac{x e^{\left (2 \, b x + 2 \, a\right )}}{c^{2} e^{\left (2 \, b x + 2 \, a\right )} +{\left (c^{2} e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (\frac{1}{2} \, \log \left (c e^{\left (b x + a\right )} + 1\right ) + \frac{1}{2} \, \log \left (-c e^{\left (b x + a\right )} + 1\right )\right )} - 1}\,{d x} - \frac{1}{2} \, b x^{2} -{\left (a + \log \left (c\right )\right )} x + x \log \left (\sqrt{c e^{\left (b x + a\right )} + 1} \sqrt{-c e^{\left (b x + a\right )} + 1} + 1\right ) - \frac{b x \log \left (c e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (-c e^{\left (b x + a\right )}\right )}{2 \, b} - \frac{b x \log \left (-c e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (c e^{\left (b x + a\right )}\right )}{2 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{asech}{\left (c e^{a + b x} \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{arsech}\left (c e^{\left (b x + a\right )}\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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